# Why is angular momentum conserved in a satellite?

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#### Explanation

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Mar 8, 2018

The principle of conservation of momentum, both linear (also known as translational) and angular, is a universal principle of Physics.

#### Explanation:

Angular momentum, $L$, is defined, in mathematical terms as

$L = I \cdot \omega$

Take the time derivative of that

$\frac{\mathrm{dL}}{\mathrm{dt}} = I \cdot \frac{\mathrm{do} m e g a}{\mathrm{dt}}$

We give the term "torque" to $I \cdot \frac{\mathrm{do} m e g a}{\mathrm{dt}}$ and we give torque the variable name $\tau$.
We give $\frac{\mathrm{do} m e g a}{\mathrm{dt}}$ the variable name $\alpha$.

Expanding that last equation with the variable names $\alpha \mathmr{and} \tau$

$\frac{\mathrm{dL}}{\mathrm{dt}} = I \cdot \frac{\mathrm{do} m e g a}{\mathrm{dt}} = I \cdot \alpha = \tau$ ... Equation1

The equation $\tau = I \cdot \alpha$ is the angular equivalent of $F = m \cdot a$ which Newton made famous in Newton's 2nd Law.

$\tau , I , \mathmr{and} \alpha$ are the angular equivalents of $F , m , \mathmr{and} a$ respectively.

If there are no external torques acting on a body, $\tau = 0$, and that there is no angular acceleration follows. That is shown in the repeat of Equation 1 below.

$\frac{\mathrm{dL}}{\mathrm{dt}} = I \cdot \frac{\mathrm{do} m e g a}{\mathrm{dt}} = I \cdot \alpha = 0$

Therefore $\frac{\mathrm{dL}}{\mathrm{dt}} = 0$. And therefore, the angular momentum is constant if $\tau = 0$.

Therefore, we have Conservation of Angular Momentum .

I hope this helps,
Steve

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vrma Share
Mar 8, 2018

The angular momentum of a satellite is its moment of momentum taken about the axis of rotation and it remains conserved as its speed remains same.

#### Explanation:

Suppose a body is moving around another mass in a circular path , its velocity changes in direction but

magnitude of the velocity remains the same in a particular orbit.

The satellite's momentum=mass x velocity =(m. v ) is always in the direction of tangent to the path

and the radius vector is drawn from the axis of rotation to the satellite;

therefore the radius vector is always normal to the direction of momentum;

So, The angular momentum = moment of linear momentum

equals to ( r X p ) a cross product of radius vector with momentum

which equals to r.p. sin 90 = r.m.v as ( sin 90)=1

as r (radius) m(mass) and v (velocity) the three are scalar values and remains the same ,

the angular momentum is constant and the direction of the angular momentum is fixed as the axis perpendicular to both r and p using

cork screw rule of vector products and is always normal to the plane of motion.

If however if any force disturbs its motion as to change its speed or path its angular momentum will no longer remain constant.

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